An inductor of 2 H and a resitance of `10Omega` are conncts in series with a bttery of 5 V. The intial rate of change of current is

Options

A. `0.5As^(-1)`
B. `2.0As^(-1)`
C. `2.5As^(-1)`
D. `0.25As^(-1)`

Answer

C

Explanation

Growth of current in the circuit is
`i=i_(0)(1-e^(-Rt//L))`
`rArr" "(di)/(dt)=(d)/(dt)i_(0)-(d)/(dt)(i_(0)e^(-Rt//L))`
`therefore" "(di)/(dt)=0-i_(0)(-(R)/(L))e^(-Rt//L)=(i_(0)R)/(L)e^(-Rt//L)`
Initially, t = 0,
`therefore" "(di)/(dt)=(i_(0)R)/(L)=(E)/(L)=(5)/(2)=2.5As^(-1)`